Another Fibonacci series

Calculus Level 3

Let F n F_n be the n n th Fibonacci number, where F 1 = 1 F_1=1 , F 2 = 1 F_2=1 and F n + 1 = F n + F n − 1 F_{n+1}=F_n+F_{n-1} for n ≥ 2 n \ge 2 . Evaluate the sum ∑ n = 2 ∞ 1 F n + 1 F n − 1 . \sum_{n=2}^\infty \frac{1}{F_{n+1} F_{n-1}}.

Hint: If you're stuck you may want to try this problem first.


The answer is 1.

Santi Spadaro by Santi Spadaro , 39, Italy

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1 solution

Chew-Seong Cheong
Aug 12, 2018

S = ∑ n = 2 ∞ 1 F n + 1 F n − 1 = ∑ n = 2 ∞ F n F n ( F n + F n − 1 ) F n − 1 = ∑ n = 2 ∞ 1 F n ( 1 F n − 1 − 1 F n + F n − 1 ) = ∑ n = 2 ∞ 1 F n ( 1 F n − 1 − 1 F n + 1 ) = ∑ n = 2 ∞ ( 1 F n − 1 F n − 1 F n F n + 1 ) = 1 F 1 F 2 − 1 F 2 F 3 + 1 F 2 F 3 − 1 F 3 F 4 + 1 F 3 F 4 − 1 F 4 F 5 + 1 F 4 F 5 − ⋯ = 1 F 1 F 2 = 1 \begin{aligned} S & = \sum_{n=2}^\infty \frac 1{{\color{#3D99F6}F_{n+1}}F_{n-1}} \\ & = \sum_{n=2}^\infty \frac {\color{#D61F06}F_n}{{\color{#D61F06}F_n}{\color{#3D99F6}\left(F_n+F_{n-1}\right)}F_{n-1}} \\ & = \sum_{n=2}^\infty \frac 1{F_n} \left(\frac 1{F_{n-1}}-\frac 1{F_n+F_{n-1}}\right) \\ & = \sum_{n=2}^\infty \frac 1{F_n} \left(\frac 1{F_{n-1}}-\frac 1{F_{n+1}}\right) \\ & = \sum_{n=2}^\infty \left(\frac 1{F_{n-1}F_n}-\frac 1{F_nF_{n+1}}\right) \\ & = \frac 1{F_1F_2} \color{#3D99F6} \cancel{- \frac 1{F_2F_3} + \frac 1{F_2F_3}} \color{#D61F06}\cancel{- \frac 1{F_3F_4} + \frac 1{F_3F_4}}\color{#3D99F6} \cancel{- \frac 1{F_4F_5} +\frac 1{F_4F_5}} - \cdots \\ & = \frac 1{F_1F_2} = \boxed 1 \end{aligned}

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